The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is

$$\begin{gathered} For constructive interference, d\sin \theta =n\lambda \\ Given , d=2\lambda \\ \Rightarrow \sin \theta =\frac{n}{2} \\ n=0,1,-1,2,-2 \\ Hence, five maxima are possible. \\ alternate solution: \\ when d=n\lambda then the number of interference maxima are given by \\ N=2n+1 \end{gathered}$$